2_Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
给你两个非空链表，代表两个整形数值，数值中的数字被反序装在链表中，链表的每个结点装了一个数字，本题要你返回这两个数值相加后的结果，结果也反序装在链表中

You may assume the two numbers do not contain any leading zero, except the number 0 itself.
你可以假设这两个数值不是以0开头，也就是假设这两个数值是正常合法的

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

我的解：

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def addTwoNumbers(self, l1, l2):        """        :type l1: ListNode        :type l2: ListNode        :rtype: ListNode        """        p = 0        result = None        current = None        while True:            addValue = l1.val + l2.val + p            p = 0            if addValue >= 10:                addValue = addValue - 10                p = 1            node = ListNode(addValue)            if result == None:                result = node                current = node            else:                current.next = node                current = node            l1 = l1.next            l2 = l2.next            if l1 == None or l2 == None:                break        if l1 == None and l2 == None:            pass        elif l1 == None or l2 == None:            rest = None            if l1 == None:                rest = l2            else :                rest = l1            while True:                addValue = p + rest.val                p = 0                if addValue >= 10:                    addValue = addValue - 10                    p = 1                node = ListNode(addValue)                current.next = node                current = node                rest = rest.next                if rest == None:                    break        if p == 0:            return result        else:            node = ListNode(p)            current.next = node            return result

看了答案才知道可以更简洁一些的 ~_~